The function f(x) defined by

$f(x)=\left\{\begin{array}{cc}\left\{x^2+e^{\frac{1}{2-x}}\right\}^{-1}, & x \neq 2 \\ \quad ~~~k \quad~~~~~~~~~~~, & x=2\end{array}\right.$

is continuous from right at x = 2, then k is equal to

$\frac{1}{4}$

It is given that f(x) is right continuous at x = 2

∴ $\lim\limits_{x \rightarrow 2^{+}} f(x)=f(2)$

$\Rightarrow \lim\limits_{h \rightarrow 0} f(2+h)=k$

$\Rightarrow \lim\limits_{h \rightarrow 0}\left\{(2+h)^2+e^{\frac{1}{2-(2+h)}}\right\}^{-1}=k$

$\Rightarrow \lim\limits_{h \rightarrow 0}\left\{(2+h)^2+e^{-1 / h}\right\}^{-1}=k$

$\Rightarrow (4+0)^{-1}=k \Rightarrow k=\frac{1}{4}$